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#circular motion

86 public questions tagged with this topic.

Solenoid, Toroid and Ampere's Law

A proton moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a uniform magnetic field of \( 0.5 \,

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Radius r = (mv/qB) . Substitute: r = (1.67 × 10⁻²⁷ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.34 × 10⁻²¹/8 × 10⁻²⁰) = 4.175 × 10⁻² m = 4.18 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

Why does a charged particle describe a circular path when moving perpendicular to a uniform magnetic field?

**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. The magnetic force F = q (v × B) acts perpendicular to the velocity, providing the centripetal force needed for circular motion when the velocity is perpendicular to the field. Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Forced Oscillations and Resonance

A particle’s x-projection from circular motion is \( x = 8 \cos (\pi t) \) (in m). What is its maximum speed?

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. Maximum speed: vₘₐₓ = ω A . A = 8 m, ω = π s⁻¹ . vₘₐₓ = π × 8 ≈ 3.14 × 8 ≈ 25.12 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 25.12 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Damped Oscillations

A particle’s x-projection from circular motion is \( x = 5 \cos (4t) \) (in m). What is its maximum speed?

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. Maximum speed: vₘₐₓ = ω A . A = 5 m, ω = 4 s⁻¹ . vₘₐₓ = 4 × 5 = 20 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A particle’s x-projection from circular motion is \( x = 6 \cos (3t) \) (in m). What is its maximum acceleration?

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. Maximum acceleration: aₘₐₓ = ω² A . A = 6 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 6 = 9 × 6 = 54 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 54 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Equations of SHM, Phase and Angular Frequency

Which statement best explains why uniform circular motion is not considered oscillatory despite being periodic?

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Oscillatory motion requires to-and-fro movement about an equilibrium, while uniform circular motion involves continuous rotation without reversing direction. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It does not involve to-and-fro motion follows, reflecting SHM depen

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Energy in SHM - Kinetic, Potential and Total

A particle’s x-projection from circular motion is \( x = 9 \cos (\pi t) \) (in m). What is its maximum speed?

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Maximum speed: vₘₐₓ = ω A . A = 9 m, ω = π s⁻¹ . vₘₐₓ = π × 9 ≈ 28.26 m/s . Applying x = A cos(ωt + φ), v =

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Periodic Motion and SHM Basic Concepts

A particle’s x-projection from circular motion is \( x = 7 \cos (3t) \) (in m). What is its maximum acceleration?

**Periodic motion** repeats after fixed period T, x(t+T)=x(t), while oscillatory motion involves to-and-fro about equilibrium. SHM is special periodic motion where restoring force proportional to displacement, F = -k x, acceleration a = -ω² x, leading to sinusoidal displacement x = A cos(ωt + φ). Maximum acceleration: aₘₐₓ = ω² A . A = 7 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 7 = 9 × 7 = 63 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 63 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Periodic Motion and SHM Basic Concepts

A particle’s x-projection from circular motion is \( x = 5 \cos (2t) \) (in m). What is its maximum speed?

**Distinction between periodic and oscillatory** clarifies all SHM is periodic but not all periodic is SHM. SHM requires linear restoring force and inertia, a ∝ -x, with ω = √(k/m). Functions like sin²ωt have period π/ω but lack a = -ω² x, thus periodic not SHM, while uniform circular motion is periodic without linear oscillation. Maximum speed: vₘₐₓ = ω A . A = 5 m, ω = 2 s⁻¹ . vₘₐₓ = 2 × 5 = 10 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 10 m/s

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Displacement, Velocity and Acceleration in SHM

A particle’s x-projection from circular motion is \( x = 8 \cos (2\pi t) \) (in m). What is its maximum acceleration?

**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Maximum acceleration: aₘₐₓ = ω² A . A = 8 m, ω = 2π s⁻¹ . aₘₐₓ = (2π)² × 8 ≈ 39.48 × 8 ≈ 315.84 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Displacement, Velocity and Acceleration in SHM

A particle in circular motion has its x-projection as \( x = 7 \cos (\pi t) \) (in m). What is its maximum speed?

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Maximum speed: vₘₐₓ = ω A . A = 7 m, ω = π s⁻¹ . vₘₐₓ = π × 7 ≈ 3.14 × 7 ≈ 21.98 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20.0 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Displacement, Velocity and Acceleration in SHM

The projection of a particle in circular motion on the x-axis is \( x = 3 \cos (2\pi t) \) (in m). What is the radius of

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. For SHM as projection of circular motion, radius = amplitude. Here, x = A cos (ω t) , so A = 3 m . Radius = 3 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.0 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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